Transitive relation

Step 1: First, ((1,2)) and ((2,3)) require ((1,3)), which is present. Step 2: Now ((1,3)) and ((3,4)) require ((1,4)), but it is missing. Step 3: Include newly implied pairs in further checking.

Step 1: ((1,2)) and ((2,3)) require ((1,3)), which is present. Step 2: ((2,3)) and ((3,4)) require ((2,4)), and ((1,3)) and ((3,4)) require ((1,4)); both are present. Step 3: Tick all required pairs in a longer chain.

Step 1: To break transitivity, two linked pairs must exist. Step 2: If the required third pair is absent, that is a counterexample. Step 3: One valid counterexample is enough to disprove a statement.

Step 1: To test transitivity, both ((a,b)) and ((b,c)) are needed. Step 2: A single pair ((2,3)) does not form such a chain, so no condition fails. Step 3: Without a counter-chain, the relation may be transitive.

Step 1: ((3,1)) and ((1,2)) have the common middle element (1). Step 2: Transitivity requires ((3,2)), but it is not present. Step 3: Try combining pairs in different valid orders, not only the listed order.

Step 1: ((3,1)) and ((1,2)) require ((3,2)). Step 2: ((3,2)) is present, and no other missing requirement occurs. Step 3: A small relation can be transitive if all required pairs are included.

Step 1: If (a\ge b) and (b\ge c), then (a\ge c) must hold. Step 2: Therefore the greater-than-or-equal relation is transitive. Step 3: Do not change the direction of the inequality.

Step 1: If (a>b) and (b>c), then (a>c). Step 2: So the (>) relation is transitive. Step 3: For both (>) and (<), use the direction of order to conclude.

Step 1: In ((1,2)) and ((2,1)), the middle element is (2). Step 2: By transitivity, the first element of the first pair and the second element of the second pair form ((1,1)). Step 3: For opposite pairs, check both directions separately.

Step 1: If student (A) is in the same class as (B), and (B) is in the same class as (C), then (A) is in the same class as (C). Step 2: So being in the same class is transitive. Step 3: Even in daily examples, test the chain condition.

Step 1: ((1,2)) and ((2,3)) require ((1,3)). Step 2: ((1,3)) is present; ((1,1)) and ((3,3)) do not create a missing required pair. Step 3: Check the main chain first, then self-pairs.

Step 1: ((1,2)) and ((2,3)) require ((1,3)). Step 2: ((1,3)) is missing, so the relation is not transitive. Step 3: The presence of ((3,3)) does not fix this missing pair.

Step 1: From ((a,b)) and ((b,c)), we get ((a,c)). Step 2: Then from ((a,c)) and ((c,d)), we get ((a,d)). Step 3: Repeat transitivity step by step in a long chain.

Step 1: Reflexivity requires ((a,a)), and symmetry requires reverse pairs. Step 2: Transitivity is a separate condition and must be checked separately. Step 3: In exams, prove each property by its own rule.

Step 1: ((5,5)) and ((5,5)) require ((5,5)) again. Step 2: This pair is present, so the condition holds. Step 3: A relation with a single self-pair can be transitive.

Step 1: ((2,1)) and ((1,3)) require ((2,3)), which is present. Step 2: ((1,2)) and ((2,1)) require ((1,1)), and ((2,1)) with ((1,2)) requires ((2,2)). Both are present. Step 3: After checking all chains, the relation is actually transitive, so option A is the trap.

Step 1: In transitivity, we only check whether every required ((a,c)) is present. Step 2: If no required pair is missing, the relation is transitive. Step 3: Do not get confused by many pairs; look for missing pairs.

Step 1: ((2,1)) and ((1,2)) require ((2,2)). Step 2: ((1,2)) and ((2,1)) require ((1,1)), which is already present. Step 3: So the minimum new pair is ((2,2)).

Step 1: ((1,2)) and ((2,4)) imply ((1,4)). Step 2: Then ((1,4)) and ((4,8)) imply ((1,8)). Step 3: Apply transitivity repeatedly along the chain.

Step 1: Transitivity is not checked by counting pairs; it is checked by finding linked pairs. Step 2: For every ((a,b)) and ((b,c)), ((a,c)) must be present. Step 3: Making a small table reduces mistakes in exams.